$f:(0,\infty) \rightarrow \mathbb{R}$ satisfies $f(x)-f(y)=f\left(\dfrac{x}{y}\right)$ for all $x,y \in (0,\infty)$ and $f(1)=0.$

Question

Suppose that $f:(0,\infty) \rightarrow \mathbb{R}$ satisfies $f(x)-f(y)=f\left(\dfrac{x}{y}\right)$ for all $x,y \in (0,\infty)$ and $f(1)=0.$

$(a)$. Prove that $f$ is continuous on $(0,\infty)$ if and only if $f$ is continuous at $1$

$(b)$. Prove that $f$ is differentiable on $(0,\infty)$ if and only if $f$ is differentiable at $1$

$(c).$ Prove that $f$ is differentiable at $1$, then $f^{\prime}(x)=\dfrac{f^{\prime}(1)}{x}$ for all $x\in (0,\infty)$.

$\underline {Attempt}$

$(a).$ If $f$ is continuous on $(0,\infty)$ then $f$ is also continuous at $1$.

$\space$ $ \space $$\space$ $\space$ If $f$ is continuous at $1,$ We have $$ \forall \space \varepsilon \space \exists \space \delta \space\text{such that} \space|x-1|<\delta \space \text{whenever} \space|f(x)-f(1)|=|f(x)|< \varepsilon $$

$\space$ $ \space $$\space$ $\space$ Now let $a \in (0,\infty)$ and if $|x-a|<\delta_1 $,

$$|f(x)-f(a)|=\left|f\left(\dfrac{x}{a}\right)\right| <\varepsilon$$

$\therefore$ $f$ is continuous on $(0,\infty)$ if and only if $f$ is continuous at $1$

$(b).$ If $f$ is differentiable $(0,\infty)$ then $f$ is also differentiable at $1$.

$\space$ $ \space $$\space$ $\space$ If $f$ is differentiable at $1$, We have $$f^{\prime}(1)=\lim_{h\to 0 } \frac{f(1+h)-f(1)}{h}$$

$\space$ $ \space $$\space$ $\space$ set $h=\dfrac{t}{x}$ implies when $h \to 0$, $t \to 0$ so $$f^{\prime}(1)=\lim_{h\to 0 } \frac{f(1+h)}{h}=\lim_{t\to 0 } \frac{f\left(1+\dfrac{t}{x}\right)}{\dfrac{t}{x}}=x\lim_{t\to 0} \frac{f(x+t)-f(x)}{t} $$

$\therefore$ $f$ is differentiable on $(0,\infty)$ if and only if $f$ is differentiable at $1$

$(c).$ If $f$ is differentiable at $1$,We have $$f^{\prime}(1)=\lim_{h\to 0 } \frac{f(1+h)-f(1)}{h}$$ $\space$ $ \space $$\space$ $\space$ this implies, $$f^{\prime}(1)=\lim_{h\to 0 } \frac{f(1+h)}{h}=x\lim_{t\to 0} \frac{f(x+t)-f(x)}{t}=xf^{\prime}(x) $$

I referred Functional equation $f(xy)=f(x)+f(y)$ and differentiability and Show a function for which $f(x + y) = f(x) + f(y) $ is continuous at zero if and only if it is continuous on $\mathbb R$ but I don't know exactly my attempt is correct or not if not give some advises.

Thank you!

Answer 1

The OP is clearly about whether the solution provided is ok, and not about a solution, I address the former.

Answer to OP: no!!

Because you neither specify what $\delta_1$ is, nor do you prove, by other means, that such a $\delta_1$ exists. It was fixed by @Paramanand Sigh in the comments. Another possible addition to OP will do the job. Add: For a given $a>0$ there is a $\delta_1>0$ such that $$ |x-a|<\delta_1 \implies |\frac{x}{a}-1|<\delta \, . $$

To emphasize: the fact that a $\delta$ exists does not automatically prove a $\delta_1$ exists. Notice that $\delta_1$ must depend on the point where you consider continuity.

Advice: When writing a proof, assume that a computer will read it. So, the first thing you do is to make sure every parameter you introduce is clearly defined either beforehand, or immediately after a comma.

Answer 2

a) I would have used the characterisation of continuity with sequences.

If $f$ is continuous in $1$, then for $x\geqslant 0$, if $x_n\rightarrow x$, then $f(x_n) = f(x)+f(\frac{x_n}{x})\rightarrow f(x) + f(1) = f(x)$. So $f$ is continuous in $x$.

b) If $f$ is differentiable on $1$, then for $x_0\geqslant 0$ and $x\neq x_0$, $\frac{f(x)-f(x_0)}{x-x_0}=\frac{1}{x_0}\cdot \frac{f(\frac{x}{x_0})-f(1)}{\frac{x}{x_0}-1}\underset{x\rightarrow x_0}{\longrightarrow}\frac{1}{x_0}f'(1)$ so $f$ is differentiable on $x$.

c) If you derive the relation with $y$ as a variable, then you get $$-f'(y) = -\frac{x}{y^2} f'(\frac{x}{y})$$ so by replacing $y$ with $1$, you get $-xf'(x)=-f'(1)$, so $$f'(x) = \frac{f'(1)}{x}$$